Subspaces of Rⁿ, Dimension, and Rank

Sets of vectors closed under the two linear operations, addition and scalar multiplication, recur throughout the subject: the span of a set of vectors, the outputs a matrix can produce, the solutions of a homogeneous system. Such a set is a subspace. Every matrix carries two of them, and basis, dimension, and rank measure their size. The general abstract vector space generalizes the construction; here everything lives inside ${\mathbb{R}}^n$.

Subspaces

A subspace is closed under linear combinations. The span of any set of vectors is the standard example: for $v_1,\dots ,v_p$ in ${\mathbb{R}}^n$, the set $\mathrm{Span}\{v_1,\dots ,v_p\}$ contains $0$, and a sum or scalar multiple of linear combinations is again a linear combination, so all three properties hold. Geometrically, $\mathrm{Span}\{v_1\}$ with $v_1\ne 0$ is a line through the origin, and $\mathrm{Span}\{v_1,v_2\}$ with $v_1,v_2$ independent is a plane through the origin.

A line not through the origin is not a subspace: it fails the first property and is closed under neither operation. The whole space ${\mathbb{R}}^n$ is a subspace of itself, and the set $\{0\}$ containing only the zero vector is the zero subspace.

Column space and null space

Two subspaces attach to every matrix, and they capture the two basic questions about $Ax=b$: which $b$ are reachable, and which $x$ collapse to zero.

Because $Ax$ is a combination of the columns, $b$ lies in $\mathrm{Col}A$ exactly when $Ax=b$ is consistent. The column space is Span of the columns, so it is described explicitly — its vectors are built directly from the columns. It equals all of ${\mathbb{R}}^m$ only when the columns span ${\mathbb{R}}^m$.

The proof checks the three properties: $A0=0$, so $0\in \mathrm{Nul}A$; if $Au=0$ and $Av=0$, then $A(u+v)=0$ and $A(cu)=0$. The null space is described implicitly — a vector belongs only if it passes the test $Ax=0$ — so producing its members takes work. Solving $Ax=0$ and writing the answer in parametric vector form turns the implicit description into an explicit spanning set.

Basis

A subspace holds infinitely many vectors, but a finite, non-redundant spanning set describes it completely.

Independence removes redundancy; spanning guarantees coverage. The columns of an invertible $n\times n$ matrix form a basis for ${\mathbb{R}}^n$, and the simplest case is the standard basis $\{e_1,\dots ,e_n\}$, the columns of $I_n$.

A basis for the null space

The parametric vector form of the solution set of $Ax=0$ yields a basis for $\mathrm{Nul}A$ directly: one basis vector per free variable.

A basis for the column space

For the column space, take the pivot columns of $A$ itself, not of an echelon form.

Row reduction changes the columns, yet it preserves every linear dependence relation among them: $Ax=0$ and $Bx=0$ have the same solutions when $B$ is an echelon form of $A$. So the non-pivot columns of $A$ are the same combinations of pivot columns that they are in $B$, hence redundant, while the pivot columns inherit the independence of the pivot columns of $B$.

Coordinate systems

The point of choosing a basis, rather than any spanning set, is uniqueness of representation.

Uniqueness follows from independence: two representations subtract to a dependence relation, which must be trivial. The unique weights are the coordinates of $x$ relative to $\mathcal{B}$, collected into the coordinate vector

The map $x\mapsto [x{]}_{\mathcal{B}}$ is a one-to-one correspondence between $H$ and ${\mathbb{R}}^p$ that preserves linear combinations — an isomorphism. A $p$-dimensional subspace of ${\mathbb{R}}^n$, whatever the ambient $n$, looks and behaves like ${\mathbb{R}}^p$ through its coordinate map; coordinate systems develop this in the abstract setting.

Dimension and rank

Every basis of a given subspace has the same number of vectors, so counting them is well defined.

Thus $\dim {\mathbb{R}}^n=n$, a plane through $0$ in ${\mathbb{R}}^3$ has dimension $2$, and a line through $0$ has dimension $1$. The dimension of $\mathrm{Nul}A$ is the number of free variables in $Ax=0$, since each free variable contributes one basis vector.

Every column of $A$ is either a pivot column, contributing to the rank, or a non-pivot column, contributing a free variable and so a dimension of the null space. Since the columns are exhausted by these two kinds, their counts add to the number of columns.

A useful shortcut follows once the dimension of a subspace is known: it already determines how many vectors a basis needs.

With the right count, only one of independent and spanning needs checking; the other follows.

The Invertible Matrix Theorem, continued

For a square matrix these dimension counts become new equivalent conditions for invertibility, extending the earlier list.¹

By the Rank Theorem, full rank $\mathrm{rank}A=n$ forces $\dim \mathrm{Nul}A=0$, so $Ax=0$ has only the trivial solution — exactly the earlier condition (d). An invertible matrix has the largest possible column space and the smallest possible null space.

Numerical note

Counting pivots to find rank is exact only with exact arithmetic. In floating point, a matrix entry stored inexactly can make a would-be zero appear nonzero (or the reverse), changing the apparent rank. The reliable notion in practice is the effective rank, read from the singular value decomposition: count the singular values that are large relative to the rounding level, rather than counting pivots.